Shape Understanding Using the Laplace Beltrami ... - Visionday

Shape Understanding Using theLaplace Beltrami OperatorKatarzyna GebalDTU Informatics

Vectors Find the basis: b 1 , b 2 ,…,b nf if ,e if i f i e i Orthonormal To find the coordinates project onto the basisvectors Meaningful We can extract/remove some information

More dimensions Discreterepresentation Scalar productg ,hi 1 g x i h x iarea i

Functions Continuous functions Scalar productorthogonalitybg x h x dxg ,ha

Fourier analysishttp://mathworld.wolfram.com/FourierSeries.html Fourier seriesb nxe inx2f nf ,b nf xnf n b nx

The 1D case – Fourier & eigenvaluesd 2dx 2 einxn 2 e inx e inx – an eigenvalue of the second derivative Linear operators on functions Map functions to functions Matrix representation in discrete case with n×nentries

The 2D case – Fourier & eigenvalues In 2 dimensions second derivative becomes:d 2d 2dx 2 dy 2 Eigenvalues can be found analytically An example – rectangles, zero at the boundarya = 1b = 1a = 2b = 1

2sin(x π) sin(1/2 y π)2sin(x π) sin(y π)2sin(x π) sin(3/2 y π)2sin(2 x π) sin(1/2 y π)2sin(x π) sin(2 y π)1.51.51.51.51.5y1y1y1y1y10.50.50.50.50.500 0.5 1x00 0.5 1x00 0.5 1x00 0.5 1x00 0.5 1x2sin(2 x π) sin(y π)2sin(x π) sin(5/2 y π)2sin(2 x π) sin(3/2 y π)2sin(2 x π) sin(2 y π)2sin(3 x π) sin(1/2 y π)1.51.51.51.51.5y1y1y1y1y10.50.50.50.50.500 0.5 1x00 0.5 1x00 0.5 1x00 0.5 1x00 0.5 1x

The Laplace-Beltrami operator The generalization of the second derivative The divergence of the gradient In many physical phenomena the heat equation the wave equationffhttp://en.wikipedia.org/wiki/File:Spherical_wave2.gif http://en.wikipedia.org/wiki/File:Heat_eqn.gif

Eigensolutions of the LBO Provide orthonormal basis The generalization of Fourier series Provide not only the basis for the function butalso a good shape descriptors that dependson the shape – eigenvalues and eigenvectorsMinimization of the gradient

LBO for a discrete meshjα in×n matrix with entries:Cotan weightsiβ icot i cot jii j i2a iif j j ijcot i cot j2a ia iotherwise ij 0Nonzero elements only in a one ringSparse matrices if we need only a small numbers ofeigenvalues

The result

Shape filtering (Bruno Levy) Our functions x, y, zpositions of the points in the space Projecto onto eigenvalues Scale or cut some coefficients Reconstruct back the shapef if ,e if i f i e ix y z